The order axioms of real numbers state
1) Either $x = y$ or $x < y$ or $x > y;$
2) If $x < y,$ then $x+z < y+z;$
3) If $x, y > 0,$ then $xy > 0;$
4) If $x > y$ and $y > z,$ then $x > z.$
To prove that we cannot give a definition of $<$ for complex numbers such that it satisfies 1), 2), and 3), note that since $i \neq 0,$ by 1) we have either $i > 0$ or $i < 0.$ Assume first that $i > 0.$ Then by 3) we have $-1 > 0.$ Adding 1 by 2) to both sides we have $0 > 1.$ On the other hand, by 3) we have $(-1)(-1) = 1 > 0,$ contradicting 1). In a similar fashion we can prove for $i < 0.$
However, why was $-1 > 0$ not already a contradiction? This is a question raised in Apostol's analysis (2nd edition, pp. 19).
$-1 > 0$ is not a contradiction by itself. You're attempting to construct a new ordering on the complex numbers, and it is not required that this ordering has any relation to the known ordering on $R$. In particular, it is not required that $-1 >0$ in this ordering.